Unique Paths - LeetCode

A robot is located at the top-left corner of a m x n grid (marked ‘Start‘ in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish‘ in the diagram below).

How many possible unique paths are there?

Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

思路：这道题有两种方法。

第一种是动态规划，构建一个m * n的数组dp, 然后dp[i][j] = dp[i - 1][j] + dp[i][j - 1]。

第二种是公式法，因为一共向下走的操作会有n - 1次，向右走的操作会有m - 1次，因此一共会走m + n - 2次。

然后从m + n - 2次操作中我们要选n - 1次向右走，因此这是一个组合问题。

排列公式 (n, m) = n! / (m!(n - m)!)

实际计算中，我们可以进一步整理为 (n - m + 1) * (n - m + 2) *...* (n - m + m) / m !

这里贴一下动态规划的代码

 1 class Solution {
 2 public:
 3     int uniquePaths(int m, int n) {
 4         vector<int> tem(n, 1);
 5         vector<vector<int> > dp(m, tem);
 6         for (int i = 1; i < m; i++)
 7             for (int j = 1; j < n; j++)
 8                 dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
 9         return dp[m - 1][n - 1];
10     }
11 };